C-使用malloc将令牌复制到2D数组

Question

努力将令牌移至2D数组。这个想法是我正在阅读具有多行的文件，获取行数，然后基于创建一个2D数组，以明智地使用内存（我不想无缘无故地创建100 x 3数组）。

我认为我在单独的函数中初始化了2D数组，但是当我尝试从strtok（）读取数据时，我会遇到错误：

error: 'arr' undeclared (first use in this function)
               strcpy(&arr[s2][c2],token);

这是我的代码：

#include <stdio.h>
#include <string.h>
int ch, lines;
int no_of_lines(char* fp)
{
    while(!feof(fp)) {
        ch = fgetc(fp);
        if(ch == 'n') {
            lines++;
        }
    }
    lines++;
    return lines;
}
void declare_space_array(int size)
{
    char* arr = (char*)malloc(size * 3 * sizeof(char));
    return;
}
int main(void)
{
    int c2 = 0;
    int s2 = 0;
    int len;
    // char data[10][4];
    static const char filename[] = "C:\Users\PC\Documents\Assignments\stringops\test.txt";
    FILE* file = fopen(filename, "r");
    no_of_lines(file);
    printf("No of lines in file = %d", lines);
    printf("n");
    // Closing file because it was read once till the end of file
    fclose(file);
    // Opening file again to read for parsing 
    file = fopen(filename, "r");
    declare_space_array(lines);
    char* token;
    if(file != NULL) {
        char line[128];                               
        while(fgets(line, sizeof line, file) != NULL) 
        {
            len = strlen(line);
            printf("%d %s", len - 1, line);
            const char s = ",";
            token = strtok(line, ",");
            while(token != NULL) {
                strcpy(arr[s2][c2], token);
                // printf( "%sn", token );
                token = strtok(NULL, ",");
                c2++;
            }
            s2++;
        }
        fclose(file);
    } else {
        perror(filename); /* why didn't the file open? */
    }
    for(r1 = 0; r1 < lines; r1++) {
        for(c1 = 0; c1 < 3; c1++) {
            printf("%s", &arr[r1][c1]);
        }
    }
    return 0;
}

文件是这样的：

A1,B1,C1
A2,B2,C2
A3,B3,C3

预期的输出会像这样：

A1
B1
C1
A2
B2
C2
A3
B3
C3

Answer 1

在聊天中讨论等之后，您可以最终得到这样的代码。这使用一个全局变量arr，它是一个指向3 char *值数组数组的指针。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static int lines = 0;
static char *(*arr)[3] = 0; // global definition.
static int no_of_lines(FILE *fp)
{
    lines = 0;
    int ch;
    while ((ch = fgetc(fp)) != EOF)
    {
        if (ch == 'n')
            lines++;
    }
    return ++lines;     // Allow for last line possibly not having a newline
}
static void declare_space_array(int size)
{
    arr = calloc(size, 3 * sizeof(char *)); // zeroed memory allocation
    if (arr == 0)
    {
        fprintf(stderr, "Failed to allocate memoryn");
        exit(1);
    }
}
int main(void)
{
    int c2 = 0;
    int s2 = 0;
    int len;
    // char data[10][4];
    // static const char filename[] = "C:\Users\PC\Documents\Assignments\stringops\test.txt";
    const char *filename = "data";
    FILE *file = fopen(filename, "r");
    if (file == 0)
    {
        fprintf(stderr, "Failed to open file '%s' for readingn", filename);
        exit(1);
    }
    no_of_lines(file);
    printf("No of lines in file = %dn", lines);
    rewind(file);
    declare_space_array(lines);
    const char delims[] = ",n";
    char line[128];
    while (fgets(line, sizeof line, file) != NULL)
    {
        char *token;
        c2 = 0;
        len = strlen(line);
        printf("%d [%.*s]n", len - 1, len - 1, line);
        token = strtok(line, delims);
        while (token != NULL)
        {
            arr[s2][c2] = strdup(token); // copy token (from strtok) into newly allocated string.
            token = strtok(NULL, delims);
            c2++;
        }
        s2++;
    }
    fclose(file);
    for (int r1 = 0; r1 < lines; r1++)
    {
        if (arr[r1][0] != 0)
        {
            for (int c1 = 0; c1 < 3; c1++)
                printf(" %-10s", arr[r1][c1]);
            putchar('n');
        }
    }
    return 0;
}

它没有释放分配的内存 - 我很懒惰。

示例数据（请注意，名称大于2个字符，并且长度可变）：

server1,Phoenix,Windows
server2,Dallas,Linux
server-99,London,z/OS

样本输出：

No of lines in file = 4
23 [server1,Phoenix,Windows]
20 [server2,Dallas,Linux]
21 [server-99,London,z/OS]
 server1    Phoenix    Windows   
 server2    Dallas     Linux     
 server-99  London     z/OS      

"文件= 4"中的行数允许在最后一行末尾没有新线。打印循环中的代码允许最终有一个新线，因此计数是过度的。只要故障在行的第一个字段，它就会发现strdup()的内存分配。如果是未成功复制的第二或第三个字段，可能会崩溃。